मराठी

X Cos Y Dy = (Xex Log X + Ex) Dx - Mathematics

Advertisements
Advertisements

प्रश्न

x cos y dy = (xex log x + ex) dx

Advertisements

उत्तर

We have, 
\[x \cos y dy = \left( x e^x \log x + e^x \right) dx\]
\[ \Rightarrow \cos y dy = \left( e^x \log x + \frac{1}{x} e^x \right)dx\]
Integrating both sides, we get
\[\int \cos y dy = \int\left( e^x \log x + \frac{1}{x} e^x \right)dx\]
\[ \Rightarrow \sin y = \log x \int e^x dx - \int\frac{1}{x} e^x dx + \int\frac{1}{x} e^x dx\]
\[ \Rightarrow \sin y = e^x \log x + C\]
\[\text{ Hence, }\sin y = e^x \log x +\text{ C is the required solution }.\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 22: Differential Equations - Exercise 22.07 [पृष्ठ ५५]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 22 Differential Equations
Exercise 22.07 | Q 7 | पृष्ठ ५५

व्हिडिओ ट्यूटोरियलVIEW ALL [2]

संबंधित प्रश्‍न

If 1, `omega` and `omega^2` are the cube roots of unity, prove `(a + b omega + c omega^2)/(c + s omega +  b omega^2) =  omega^2`


Show that y = AeBx is a solution of the differential equation

\[\frac{d^2 y}{d x^2} = \frac{1}{y} \left( \frac{dy}{dx} \right)^2\]

Show that Ax2 + By2 = 1 is a solution of the differential equation x \[\left\{ y\frac{d^2 y}{d x^2} + \left( \frac{dy}{dx} \right)^2 \right\} = y\frac{dy}{dx}\]

 


Verify that y2 = 4a (x + a) is a solution of the differential equations
\[y\left\{ 1 - \left( \frac{dy}{dx} \right)^2 \right\} = 2x\frac{dy}{dx}\]


Verify that y = log \[\left( x + \sqrt{x^2 + a^2} \right)^2\]  satisfies the differential equation \[\left( a^2 + x^2 \right)\frac{d^2 y}{d x^2} + x\frac{dy}{dx} = 0\]


\[x\left( x^2 - 1 \right)\frac{dy}{dx} = 1, y\left( 2 \right) = 0\]

\[\frac{dy}{dx} = \left( e^x + 1 \right) y\]

\[5\frac{dy}{dx} = e^x y^4\]

\[x\frac{dy}{dx} + y = y^2\]

\[\frac{dy}{dx} = \frac{x e^x \log x + e^x}{x \cos y}\]

\[\sqrt{1 + x^2} dy + \sqrt{1 + y^2} dx = 0\]

(1 − x2) dy + xy dx = xy2 dx


\[\frac{dy}{dx} = y \tan 2x, y\left( 0 \right) = 2\] 

\[\frac{dy}{dx} = y \tan x, y\left( 0 \right) = 1\]

\[\frac{dy}{dx} = 2xy, y\left( 0 \right) = 1\]

\[\frac{dy}{dx} = 1 + x^2 + y^2 + x^2 y^2 , y\left( 0 \right) = 1\]

\[\frac{dy}{dx} = 1 + x + y^2 + x y^2\] when y = 0, x = 0

Find the particular solution of edy/dx = x + 1, given that y = 3, when x = 0.


Find the particular solution of the differential equation \[\frac{dy}{dx} = - 4x y^2\]  given that y = 1, when x = 0.


In a bank principal increases at the rate of r% per year. Find the value of r if ₹100 double itself in 10 years (loge 2 = 0.6931).


Find the particular solution of the differential equation
(1 – y2) (1 + log x) dx + 2xy dy = 0, given that y = 0 when x = 1.


x2 dy + y (x + y) dx = 0


\[\frac{dy}{dx} = \frac{x}{2y + x}\]

(x + 2y) dx − (2x − y) dy = 0


Find the equation of the curve which passes through the point (2, 2) and satisfies the differential equation
\[y - x\frac{dy}{dx} = y^2 + \frac{dy}{dx}\]


Find the equation of the curve which passes through the point (3, −4) and has the slope \[\frac{2y}{x}\]  at any point (x, y) on it.


The solution of the differential equation y1 y3 = y22 is


The differential equation satisfied by ax2 + by2 = 1 is


Integrating factor of the differential equation cos \[x\frac{dy}{dx} + y \sin x = 1\], is


The integrating factor of the differential equation \[x\frac{dy}{dx} - y = 2 x^2\]


If xmyn = (x + y)m+n, prove that \[\frac{dy}{dx} = \frac{y}{x} .\]


Form the differential equation from the relation x2 + 4y2 = 4b2


For the following differential equation find the particular solution.

`(x + 1) dy/dx − 1 = 2e^(−y)`,

when y = 0, x = 1


Solve the following differential equation.

`x^2 dy/dx = x^2 +xy - y^2`


Choose the correct alternative:

Differential equation of the function c + 4yx = 0 is


State whether the following statement is True or False:

The integrating factor of the differential equation `("d"y)/("d"x) - y` = x is e–x 


Solve: ydx – xdy = x2ydx.


Solve the differential equation `"dy"/"dx"` = 1 + x + y2 + xy2, when y = 0, x = 0.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×